# Calculus/Limits/Exercises

 ← Proofs of Some Basic Limit Rules Calculus Differentiation → Limits/Exercises

## Basic Limit Exercises

1. $\lim _{x\to 2}{\Big [}4x^{2}-3x-1{\Big ]}$
$9$
$9$
2. $\lim _{x\to 5}{\Big [}x^{2}{\Big ]}$
$25$
$25$
3. $\lim _{x\to {\frac {\pi }{4}}}{\Big [}\cos ^{2}(x){\Big ]}$
$1/2$
$1/2$
4. $\lim _{x\to 1}{\Big [}5e^{x-1}-5{\Big ]}$
$0$
$0$

## One-Sided Limits

Evaluate the following limits or state that the limit does not exist.

5. $\lim _{x\to 0^{-}}{\frac {x^{3}+x^{2}}{x^{3}+2x^{2}}}$
${\frac {1}{2}}$
${\frac {1}{2}}$
6. $\lim _{x\to 7^{-}}{\Big [}|x^{2}+x|-x{\Big ]}$
$49$
$49$
7. $\lim _{x\to -1^{+}}{\sqrt {1-x^{2}}}$
$0$
$0$
8. $\lim _{x\to -1^{-}}{\sqrt {1-x^{2}}}$
The limit does not exist.
The limit does not exist.
9. $\lim _{x\to 0^{-}}{\frac {|x|}{x}}$
$-1$
$-1$
10. $\lim _{x\to 1^{-}}\arcsin(x)$
${\frac {\pi }{2}}$
${\frac {\pi }{2}}$

## Two-Sided Limits

Evaluate the following limits or state that the limit does not exist.

11. $\lim _{x\to -1}{\frac {1}{x-1}}$
$-{\frac {1}{2}}$
$-{\frac {1}{2}}$
12. $\lim _{x\to 4}{\frac {1}{x-4}}$
The limit does not exist.
The limit does not exist.
13. $\lim _{x\to 2}{\frac {1}{x-2}}$
The limit does not exist.
The limit does not exist.
14. $\lim _{x\to -3}{\frac {x^{2}-9}{x+3}}$
$-6$
$-6$
15. $\lim _{x\to 3}{\frac {x^{2}-9}{x-3}}$
$6$
$6$
16. $\lim _{x\to -1}{\frac {x^{2}+2x+1}{x+1}}$
$0$
$0$
17. $\lim _{x\to -1}{\frac {x^{3}+1}{x+1}}$
$3$
$3$
18. $\lim _{x\to 4}{\frac {x^{2}+5x-36}{x^{2}-16}}$
${\frac {13}{8}}$
${\frac {13}{8}}$
19. $\lim _{x\to 25}{\frac {x-25}{{\sqrt {x}}-5}}$
$10$
$10$
20. $\lim _{x\to 0}{\frac {|x|}{x}}$
The limit does not exist.
The limit does not exist.
21. $\lim _{x\to 2}{\frac {1}{(x-2)^{2}}}$
$\infty$
$\infty$
22. $\lim _{x\to 3}{\frac {\sqrt {x^{2}+16}}{x-3}}$
The limit does not exist.
The limit does not exist.
23. $\lim _{x\to -2}{\frac {3x^{2}-8x-3}{2x^{2}-18}}$
$-{\frac {5}{2}}$
$-{\frac {5}{2}}$
24. $\lim _{x\to 2}{\frac {x^{2}+2x+1}{x^{2}-2x+1}}$
$9$
$9$
25. $\lim _{x\to 3}{\frac {x+3}{x^{2}-9}}$
The limit does not exist.
The limit does not exist.
26. $\lim _{x\to -1}{\frac {x+1}{x^{2}+x}}$
$-1$
$-1$
27. $\lim _{x\to 1}{\frac {1}{x^{2}+1}}$
${\frac {1}{2}}$
${\frac {1}{2}}$
28. $\lim _{x\to 1}\left[x^{2}+5x-{\frac {1}{2-x}}\right]$
$5$
$5$
29. $\lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}$
$0$
$0$
30. $\lim _{x\to 1}{\frac {x^{2}-1}{x^{2}+2x-3}}$
${\frac {1}{2}}$
${\frac {1}{2}}$
31. $\lim _{x\to 1}{\frac {5x}{x^{2}+2x-3}}$
The limit does not exist.
The limit does not exist.
32. $\lim _{x\to -1}\ln({\sqrt {x^{2}-1}})$
The limit does not exist.
The limit does not exist.
33. $\lim _{x\to 1}\arcsin(x)$
The limit does not exist.
The limit does not exist.

## Limits to Infinity

Evaluate the following limits or state that the limit does not exist.

34. $\lim _{x\to \infty }{\frac {-x+\pi }{x^{2}+3x+2}}$
$0$
$0$
35. $\lim _{x\to -\infty }{\frac {x^{2}+2x+1}{3x^{2}+1}}$
${\frac {1}{3}}$
${\frac {1}{3}}$
36. $\lim _{x\to -\infty }{\frac {3x^{2}+x}{2x^{2}-15}}$
${\frac {3}{2}}$
${\frac {3}{2}}$
37. $\lim _{x\to -\infty }{\Big [}3x^{2}-2x+1{\Big ]}$
$\infty$
$\infty$
38. $\lim _{x\to \infty }{\frac {2x^{2}-32}{x^{3}-64}}$
$0$
$0$
39. $\lim _{x\to \infty }6$
$6$
$6$
40. $\lim _{x\to \infty }{\frac {3x^{2}+4x}{x^{4}+2}}$
$0$
$0$
41. $\lim _{x\to -\infty }{\frac {2x+3x^{2}+1}{2x^{2}+3}}$
${\frac {3}{2}}$
${\frac {3}{2}}$
42. $\lim _{x\to -\infty }{\frac {x^{3}-3x^{2}+1}{3x^{2}+x+5}}$
$-\infty$
$-\infty$
43. $\lim _{x\to \infty }{\frac {x^{2}+2}{x^{3}-2}}$
$0$
$0$
44. $\lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}$
$0$
$0$
45. $\lim _{x\to -\infty }x^{2}\cos \left({\frac {1}{x}}\right)$
$\infty$
$\infty$
46. $\lim _{x\to \infty }{\frac {\sin(\arctan(x))}{\arctan(-x)}}$
$-{\frac {2}{\pi }}$
$-{\frac {2}{\pi }}$

## Limits of Piecewise Functions

Evaluate the following limits or state that the limit does not exist.

48. Consider the function

$f(x)={\begin{cases}(x-2)^{2}&{\text{if }}x<2\\x-3&{\text{if }}x\geq 2\end{cases}}$
a. $\lim _{x\to 2^{-}}f(x)$
$0$
$0$
b. $\lim _{x\to 2^{+}}f(x)$
$-1$
$-1$
c. $\lim _{x\to 2}f(x)$
The limit does not exist
The limit does not exist

49. Consider the function

$g(x)={\begin{cases}-2x+1&{\text{if }}x\leq 0\\x+1&{\text{if }}0
a. $\lim _{x\to 4^{+}}g(x)$
$18$
$18$
b. $\lim _{x\to 4^{-}}g(x)$
$5$
$5$
c. $\lim _{x\to 0^{+}}g(x)$
$1$
$1$
d. $\lim _{x\to 0^{-}}g(x)$
$1$
$1$
e. $\lim _{x\to 0}g(x)$
$1$
$1$
f. $\lim _{x\to 1}g(x)$
$2$
$2$

50. Consider the function

$h(x)={\begin{cases}2x-3&{\text{if }}x<2\\8&{\text{if }}x=2\\-x+3&{\text{if }}x>2\end{cases}}$
a. $\lim _{x\to 0}h(x)$
$-3$
$-3$
b. $\lim _{x\to 2^{-}}h(x)$
$1$
$1$
c. $\lim _{x\to 2^{+}}h(x)$
$1$
$1$
d. $\lim _{x\to 2}h(x)$
$1$
$1$

## Intermediate Value Theorem

51. Use the intermediate value theorem to show that there exists a value $f(x)=3$  for $f(x)={\sqrt {x}}+{\frac {1}{\sqrt {x}}}$  from $1\leq x\leq 9$ . If you cannot use the intermediate value theorem to show this, explain why.
Notice $f(x)$  is continuous from $x\in \left[1,9\right]$ . Ergo, the intermediate value theorem applies. For all $y\in \left[2,{\frac {10}{3}}\right]$ , there exists a $c\in (1,9)$  so that $f(c)=3$ . $\blacksquare$
Notice $f(x)$  is continuous from $x\in \left[1,9\right]$ . Ergo, the intermediate value theorem applies. For all $y\in \left[2,{\frac {10}{3}}\right]$ , there exists a $c\in (1,9)$  so that $f(c)=3$ . $\blacksquare$
52. Use the intermediate value theorem to show that there exists an $x=\mu$  so that $f(\mu )={\frac {\pi }{4}}$  for $f(x)={\frac {\sin(x)-\cos(x)}{\ln {x}}}$  from $\pi \leq x\leq {\frac {9}{4}}\pi$ . If you cannot use the intermediate value theorem to show this, explain why.
Notice $f(x)$  is continuous from $x\in \left[\pi ,{\frac {9}{4}}\pi \right]$ . Ergo, the intermediate value theorem applies.
$f(\pi )={\frac {1}{\ln(\pi )}}$
$f\left({\frac {9}{4}}\pi \right)=0$

It is known the following is true: $1\geq {\frac {\pi }{4}}\geq 0$ . From there, we can directly argue the following:

{\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}
By the intermediate value theorem, if $f(x)$  is continuous from $x\in \left[\pi ,{\frac {9}{4}}\pi \right]$ , then there exists an $x=\mu$  so that $f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )$  for $\mu \in \left[\pi ,{\frac {9}{4}}\pi \right]$ . $\blacksquare$
Notice $f(x)$  is continuous from $x\in \left[\pi ,{\frac {9}{4}}\pi \right]$ . Ergo, the intermediate value theorem applies.
$f(\pi )={\frac {1}{\ln(\pi )}}$
$f\left({\frac {9}{4}}\pi \right)=0$

It is known the following is true: $1\geq {\frac {\pi }{4}}\geq 0$ . From there, we can directly argue the following:

{\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}
By the intermediate value theorem, if $f(x)$  is continuous from $x\in \left[\pi ,{\frac {9}{4}}\pi \right]$ , then there exists an $x=\mu$  so that $f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )$  for $\mu \in \left[\pi ,{\frac {9}{4}}\pi \right]$ . $\blacksquare$
53. Use the intermediate value theorem to show that there exists a value $x=\Gamma$  so that $f(\Gamma )=2$  for $f(x)={\sqrt {-2x+4}}+5\ln \left(x^{2}\right)$  from $-1\leq x\leq 1$ . If you cannot use the intermediate value theorem to show this, explain why.
Notice $f(x)$  is not continuous for $x=0$  since $\lim _{x\to 0}f(x)$  is unbounded. Ergo, the intermediate value theorem cannot be used to solve this problem.
Notice $f(x)$  is not continuous for $x=0$  since $\lim _{x\to 0}f(x)$  is unbounded. Ergo, the intermediate value theorem cannot be used to solve this problem.